Let f :[-3,1] rightarrow R be given asf(x)=be | vedclass

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(B) Given $f(x) = \begin{cases} \min \{(x+6), x^{2}\}, & -3 \leq x \leq 0 \ \max \{\sqrt{x}, x^{2}\}, & 0 \leq x \leq 1 \end{cases}$For $-3 \leq x \l

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$41$

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(B) Given $f(x) = \begin{cases} \min \{(x+6), x^{2}\}, & -3 \leq x \leq 0 \ \max \{\sqrt{x}, x^{2}\}, & 0 \leq x \leq 1 \end{cases}$For $-3 \leq x \leq 0$,we compare $x+6$ and $x^2$. They intersect at $x^2 = x+6 \implies x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0$. Since $x \in [-3, 0]$,the intersection is at $x = -2$.For $x \in [-3, -2]$,$x^2 \geq x+6$,so $\min \{(x+6), x^2\} = x+6$.For $x \in [-2, 0]$,$x^2 \leq x+6$,so $\min \{(x+6), x^2\} = x^2$.For $0 \leq x \leq 1$,we compare $\sqrt{x}$ and $x^2$. They intersect at $\sqrt{x} = x^2 \implies x = x^4 \implies x(x^3-1) = 0$,so $x=0, 1$.For $x \in [0, 1]$,$\sqrt{x} \geq x^2$,so $\max \{\sqrt{x}, x^2\} = \sqrt{x}$.The area $A$ is given by:$A = \int_{-3}^{-2} (x+6) dx + \int_{-2}^{0} x^2 dx + \int_{0}^{1} \sqrt{x} dx$$A = \left[ \frac{x^2}{2} + 6x ight]_{-3}^{-2} + \left[ \frac{x^3}{3} ight]_{-2}^{0} + \left[ \frac{2}{3} x^{3/2} ight]_{0}^{1}$$A = \left( (2 - 12) - (4.5 - 18) ight) + \left( 0 - (-8/3) ight) + \left( 2/3 - 0 ight)$$A = (-10 + 13.5) + 8/3 + 2/3 = 3.5 + 10/3 = 7/2 + 10/3 = (21+20)/6 = 41/6$.Therefore,$6A = 6 	imes (41/6) = 41$.

Let $f :[-3,1] \rightarrow R$ be given as
$f(x)=\begin{cases} \min \{(x+6), x^{2}\}, & -3 \leq x \leq 0 \\ \max \{\sqrt{x}, x^{2}\}, & 0 \leq x \leq 1 \end{cases}$
If the area bounded by $y = f(x)$ and the $x$-axis is $A$,then the value of $6A$ is equal to ....... .

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Let $f :[-3,1] \rightarrow R$ be given as$f(x)=\begin{cases} \min \{(x+6), x^{2}\}, & -3 \leq x \leq 0 \\ \max \{\sqrt{x}, x^{2}\}, & 0 \leq x \leq 1 \end{cases}$If the area bounded by $y = f(x)$ and the $x$-axis is $A$,then the value of $6A$ is equal to ....... .